Question

question 10
0/1 pt 3 99
find the standard deviation of this probability distribution. give your answer to at least 2 decimal places

Explanation:

Step1: Calculate the mean $\mu$

$\mu=\sum_{x}x\cdot P(x)=0\times0.25 + 1\times0.05+2\times0.25 + 3\times0.45=0 + 0.05+0.5+1.35 = 1.9$

Step2: Calculate $(x - \mu)^2\cdot P(x)$ for each $x$

For $x = 0$: $(0 - 1.9)^2\times0.25=( - 1.9)^2\times0.25 = 3.61\times0.25=0.9025$
For $x = 1$: $(1 - 1.9)^2\times0.05=( - 0.9)^2\times0.05 = 0.81\times0.05 = 0.0405$
For $x = 2$: $(2 - 1.9)^2\times0.25=(0.1)^2\times0.25 = 0.01\times0.25 = 0.0025$
For $x = 3$: $(3 - 1.9)^2\times0.45=(1.1)^2\times0.45 = 1.21\times0.45 = 0.5445$

Step3: Calculate the variance $\sigma^{2}$

$\sigma^{2}=\sum_{x}(x - \mu)^2\cdot P(x)=0.9025+0.0405 + 0.0025+0.5445=1.49$

Step4: Calculate the standard - deviation $\sigma$

$\sigma=\sqrt{\sigma^{2}}=\sqrt{1.49}\approx1.22$

Answer:

$1.22$